Question

Three numbers whose sum is $15$ are in AP. If $1,4,19$ are added to them respectively, then they are in GP. How many triplet sets of such numbers are possible?

Answer 1

Hint: Here we need are given that three numbers are in AP so we will let them to be $a - d,a,a + d$ where $d$ is the common difference and $a - d,a,a + d$ are three terms in AP. So we can get their sum as $15$
So we can find $a$ by this and then we will add the numbers given to the three numbers and make them the GP and we know that when $a,b,c$ are in GP then ${b^2} = ac$ and by this we can easily get the triplets.

Complete step-by-step answer:
Here we are given that the three terms are in AP. We know that whenever we have odd number of terms in AP and their sum is given then it is very sufficient to take the term as $......a - 2d,a - d,a,a + d,a + 2d,.......$ so that when we add the terms we get the value of $a$ and our calculation is easier.
So here we are given that three terms are in AP so let the three terms be $a - d,a,a + d$
Their sum is $15$ so we can say that
$
  \Rightarrow a - d + a + a + d = 15 \\
  \Rightarrow 3a = 15 \\
  \Rightarrow a = 5 \\
 $
So we get three numbers as $5 - d,5,5 + d$
Now if we add the numbers $1,4,19$ to these we will get the numbers as $5 - d + 1,5 + 4,5 + d + 19$ which is equal to $6 - d,9,24 + d$
Now we are given that these three terms are in GP which means the geometric progression.
Now we know that when $a,b,c$ are in GP then ${b^2} = ac$
So we can say here that
$
  \Rightarrow {9^2} = (6 - d)(24 + d) \\
  \Rightarrow 81 = 6(24 + d) - d(24 + d) \\
  \Rightarrow 81 = 144 + 6d - 24d - {d^2} \\
  \Rightarrow {d^2} + 18d - 63 = 0 \\
 $
Now we need to simplify it and get
$
  \Rightarrow {d^2} + 21d - 3d - 63 = 0 \\
  \Rightarrow d(d + 21) - 3(d + 21) = 0 \\
  \Rightarrow (d - 3)(d + 21) = 0 \\
  \Rightarrow d = 3, - 21 \\
 $
So we get the two values of $d$ so we can say that there will be two triplets which will be
By putting $d = 3$ we get the three numbers as:
$\Rightarrow$ $5 - 3,5,5 + 3 = 2,5,8$
By putting $d = - 21$ we get the numbers as
$\Rightarrow$ $5 + 21,5,5 - 21 = 26,5, - 16$
So two triplets are $2,5,8{\text{ and }}26,5, - 16$

Note: Here the students can make the calculations harder by letting the three numbers as $a,b,c$ or any other different variables because when we choose different variable then we need to first solve the two equations to get $a,b,c$ while by letting them as $a - d,a,a + d$ on adding $d$ is cancelled out and we get the value of $a$ easily.